Usual speed 2

A passenger train takes 1 h less for a journey of 120 km.If its speed increased by 10 k/h from its usual speed. What is its usual speed?

Correct answer:

v = 30 km/h

Δ v = 10 km/h Δ t = 1 h s = 120 km s = v \cdot t = (v + Δ v) \cdot (t - Δ t) v \cdot t = (v + Δ v) \cdot (t - Δ t) v \cdot t = v \cdot t + Δ v \cdot t - Δ t \cdot v - Δ t \cdot Δ v 0 = Δ v \cdot t - Δ t \cdot v - Δ t \cdot Δ v Δ t \cdot v + Δ t \cdot Δ v = Δ v \cdot t Δ t \cdot v + Δ t \cdot Δ v = Δ v \cdot (s / v) Δ t \cdot v^{2} + Δ t \cdot Δ v \cdot v = Δ v \cdot s 1 \cdot v^{2} + 1 \cdot 10 \cdot v = 10 \cdot 120 v^{2} + 10 v - 1200 = 0 a = 1; b = 10; c = - 1200 D = b^{2} - 4 a c = 1 0^{2} - 4 \cdot 1 \cdot (- 1200) = 4900 D > 0 v_{1, 2} = \frac{- b \pm D}{2 a} = \frac{- 1 0 \pm 4 9 0 0}{2} v_{1, 2} = \frac{- 1 0 \pm 7 0}{2} v_{1, 2} = - 5 \pm 35 v_{1} = 30 v_{2} = - 40 v > 0 v = v_{1} = 30 km/h Verifying Solution: t_{1} = s / v = 120 / 30 = 4 h t_{2} = \frac{s}{v + Δ v} = \frac{1 2 0}{3 0 + 1 0} = 3 h

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